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Blababa [14]
3 years ago
10

A group of 25 particles have the following speeds: two have speed 11m/s , seven have 16m/s , four have 22m/s , three have 25m/s

, six have 31m/s , one has 34m/s , and two have 41 m/s .
a. Determine the average speed.
vavg= ______________m/s
b. Determine the rms speed.
vrms= ______________m/s
c. Determine the most probable speed.
Physics
1 answer:
Fittoniya [83]3 years ago
8 0

Answer:

a) vavg = 16.59 m/s

b) vrms= 26.42 m/s

c) the most probable speed is the average speed = 16.59 m/s  (since the is no particle with 16.59 m/s → 16m/s is the most probable speed observed)

Explanation:

a) the average velocity will be

vavg =( 1/total number of particles )*∑ velocity * number of particles  

= (2*11 + 4*16 + 4*22 +3*25 + 6*31  + 1*34 + 2*41)/ (2 + 4 + 4 +3 + 6 + 1 + 2) = 16.59 m/s

b) the root mean squared speed (rms) will be

vrms=√[( 1/total number of particles )*∑ velocity² * number of particles] =   √[ (2*11² + 4*16² + 4*22² +3*25² + 6*31²  + 1*34² + 2*41²)/ (2 + 4 + 4 +3 + 6 + 1 + 2)]  = 26.42 m/s

c) if any particle is chosen at random then the probability of choosing that particle will be the given be the fraction of particles over the total , and thus the expected speed (most probable speed ) will be the average speed = 16.59 m/s ( since the is no particle with 16.59 m/s → 16m/s is the most probable speed observed)

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dlinn [17]

Answer:

9.8 Joules (rounded to 2 significant figures)

Explanation:

Work done (J)= Force(N) x distance changed (m)

  • Force= 9.80665 x 0.5kg
  • Force= 4.90332 Newtons

  • Distance changed= 5-3
  • distance changed= 2m/s

--> work done= 4.90332 x 2

work done= 9.8 Joules

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2 years ago
What effect will a turning point have on an individuals life
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4 0
2 years ago
A spring with a spring constant of 50 N/m is stretched 15cm. What is the force and energy associated with this stretching?
Olenka [21]
Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m

Formula:
F = k*x

Solving: 
F = k*x
F = 50*0.15
\boxed{\boxed{F = 7.5\:N}}\end{array}}\qquad\quad\checkmark

Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m

Formula:
E = \frac{k*x^2}{2}

Solving:(Energy associated with this stretching)
E = \frac{k*x^2}{2}
E =  \frac{50*0.15^2}{2}
E =  \frac{50*0.0225}{2}
E =  \frac{1.125}{2}
\boxed{\boxed{E = 0.5625\:J}}\end{array}}\qquad\quad\checkmark

7 0
3 years ago
A 5.00-m-long uniform ladder, weighing 200 N, rests against a smooth vertical wall with its base on a horizontal rough floor, a
Morgarella [4.7K]

Answer:

0.488  m  

Explanation:

If θ be the angle ladder makes with the plane

cos θ = 1.2 / 5

Tan θ = 4.04

Let the height a person of weight 600 N  can climb be h from the ground .

Distance from the base point  where ladder touches the floor  = h / tanθ

= h / 4.04

Total reaction force = total downward force

R = 200 + 600

800 N

Frictional force = μ R

= .2 x 800

= 160 N

Taking moment of force about the point on the ladder  where it  touches the floor  and balancing them

200 x 1.2 x .5 + 600 x   h / tanθ  = μ R x  1.2 / tanθ ( reaction  at the top point of ladder where it touches the wall is  R₁ and

R₁ =μ R   )

= 200 x 1.2 x .5 + 600 x   h / tanθ  = 160 x 1.2 / tanθ

120  - 600 h / 4.04 = 47.52

120 - 47.52 = 600 h / 4.04

72.48= 148.51 h

h = 0.488  m  

=

5 0
3 years ago
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